Simplify and expand the following expression: $ \dfrac{2}{t - 2}+ \dfrac{5}{t - 9}+ \dfrac{2t}{t^2 - 11t + 18} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{2t}{t^2 - 11t + 18} = \dfrac{2t}{(t - 2)(t - 9)}$ Now we have: $ \dfrac{2}{t - 2}+ \dfrac{5}{t - 9}+ \dfrac{2t}{(t - 2)(t - 9)} $ The least common multiple of the denominators is: $ (t - 2)(t - 9)$ In order to get the first term over $(t - 2)(t - 9)$ , multiply by $\dfrac{t - 9}{t - 9}$ $ \dfrac{2}{t - 2} \times \dfrac{t - 9}{t - 9} = \dfrac{2(t - 9)}{(t - 2)(t - 9)} $ In order to get the second term over $(t - 2)(t - 9)$ , multiply by $\dfrac{t - 2}{t - 2}$ $ \dfrac{5}{t - 9} \times \dfrac{t - 2}{t - 2} = \dfrac{5(t - 2)}{(t - 2)(t - 9)} $ Now we have: $ \dfrac{2(t - 9)}{(t - 2)(t - 9)} + \dfrac{5(t - 2)}{(t - 2)(t - 9)} + \dfrac{2t}{(t - 2)(t - 9)} $ $ = \dfrac{ 2(t - 9) + 5(t - 2) + 2t} {(t - 2)(t - 9)} $ Expand: $ = \dfrac{2t - 18 + 5t - 10 + 2t}{t^2 - 11t + 18} $ $ = \dfrac{9t - 28}{t^2 - 11t + 18}$